I'm studying set theory using Halmos'book.
In the book, the definition of an ordinal is :
A set $S$ is an ordinal if and only if $S$ is strictly well-ordered with respect to set membership and every element of $S$ is also a subset of $S$.
Does the above definition imply that a set $x$ which are not $w$ (the set of all natural numbers) and its successors ($w^+$, $(w^+)^+$,…) must be a natural number ?
(i.e. does there exist, say an ordinal $cat$ or $orange$ of which every element of it is one of its subsets ? or when we speak of ordinals, it must be 0, 1, 2, …, $w$, $w^+$, $(w^+)^+$, … i.e. the natural numbers and its successors)
Thank you very much for your explanation!
Best Answer
No, there are many more ordinals -- you've only seen the tip of the iceberg! The first ordinal not on your list is usually called $\omega \cdot 2$ (read "omega times two"), and it consists exactly of the natural numbers, the set $\omega$, and all of $\omega$'s iterated successors. That is, $$ \omega \cdot 2 = \{0, 1, 2, \ldots, \omega, \omega^+, (\omega^+)^+, \ldots\}. $$ (Usually what you call $\omega^+$ is called $\omega + 1$, and $(\omega^+)^+ = \omega + 2$, and so on.) You could call $\omega \cdot 2$ cat if you want to, but $\omega \cdot 2$ is definitely a more common name for it. :-)
Presumably you can now guess at what $\omega \cdot 3$ is, and $\omega \cdot 4$, up to $\omega \cdot n$ for any natural $n$. But we're not done of course, after that you can add up all of those (and their successors) and get $\omega \cdot \omega$, or $\omega^2$. And then you might be able to guess what $\omega^2 \cdot 2$ is, and $\omega^2 \cdot 3$, and maybe $\omega^2 \cdot \omega$ -- and then the notation $\omega^3$ starts to make sense. And then $\omega^4$, and then eventually $\omega^\omega$. And still this is only a very small tip of the iceberg!
Enjoy exploring!
Edit: based on the clarification in the comments, maybe this better answers your question.
Proposition: For any ordinals $\alpha, \beta$, either $\alpha \subseteq \beta$ or $\beta \subseteq \alpha$.
Proof: Suppose towards a contradiction that this is false; that is, $\alpha$ contains elements not in $\beta$, and vice versa. Since $\alpha$ is a well-order, we can find the least element of $\alpha$ that is not in $\beta$; call it $\alpha'$. Similarly, let $\beta'$ be the least element in $\beta$ that is not in $\alpha$.
Let us show that $\alpha' = \beta'$, contradicting our assumption. It suffices to prove $\alpha' \subseteq \beta'$ and $\beta' \subseteq \alpha'$ -- WLOG let's prove the former.
Take $\alpha'' \in \alpha'$. This means that $\alpha'' \in \alpha$, and in the ordering in $\alpha$, $\alpha'' < \alpha'$, so by definition of $\alpha'$ we have that $\alpha'' \in \beta$. We cannot have that $\beta' < \alpha''$, because that would mean $\beta' \in \alpha''$ and thus $\beta' \in \alpha$. Similarly, we cannot have $\beta' = \alpha''$, because then we would also have $\beta' \in \alpha$. Thus we have $\alpha'' < \beta'$, or in other words $\alpha'' \in \beta'$. It follows that $\alpha' \subseteq \beta'$.
This is a contradiction, so $\alpha \subseteq \beta$ or $\beta \subseteq \alpha$. $\square$
That tells you, basically, that all the ordinals there are, are in one long line. Every ordinal (as a set) consists exactly of all ordinals below it, and there is no "branching". In particular, the natural numbers are the only finite ordinals, and every infinite ordinal contains the natural numbers.